SaraNextGen.Com
Updated By SaraNextGen
On March 11, 2024, 11:35 AM

Page No 55: - Chapter 2 Solutions Exercise Solutions class 12 ncert solutions Chemistry - SaraNextGen [2024]


Question 2.9:

Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering.

Answer:

It is given that vapour pressure of water, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/261/6118/NCERT(INTEXT)_18-11-08)_Utpal_12_Chemistry_2_12_html_m13ba95e9.gif = 23.8 mm of Hg

Weight of water taken, w1 = 850 g

Weight of urea taken, w2 = 50 g

Molecular weight of water, M1 = 18 g mol−1

Molecular weight of urea, M2 = 60 g mol−1

Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p1.

Now, from Raoult’s law, we have:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/261/6118/NCERT(INTEXT)_18-11-08)_Utpal_12_Chemistry_2_12_html_m4200b02a.gif

Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173.

Question 2.10:

Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C. Molal elevation constant for water is 0.52 K kg mol−1.

Answer:

Here, elevation of boiling point ΔTb = (100 + 273) − (99.63 + 273)

= 0.37 K

Mass of water, wl = 500 g

Molar mass of sucrose (C12H22O11), M2 = 11 × 12 + 22 × 1 + 11 × 16

= 342 g mol−1

Molal elevation constant, Kb = 0.52 K kg mol−1

We know that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/261/6120/NCERT(INTEXT)_18-11-08)_Utpal_12_Chemistry_2_12_html_m4ace1a61.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/261/6120/NCERT(INTEXT)_18-11-08)_Utpal_12_Chemistry_2_12_html_5e169480.gif

= 121.67 g (approximately)

Hence, 121.67 g of sucrose is to be added.

Note: There is a slight variation in this Answer and the one given in the NCERT textbook.

Question 2.11:

Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf = 3.9 K kg mol−1.

Answer:

Mass of acetic acid, w1 = 75 g

Molar mass of ascorbic acid (C6H8O6), M2 = 6 × 12 + 8 × 1 + 6 × 16

= 176 g mol−1

Lowering of melting point, ΔTf = 1.5 K

We know that:

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/261/6123/NCERT(INTEXT)_18-11-08)_Utpal_12_Chemistry_2_12_html_7a98b398.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/261/6123/NCERT(INTEXT)_18-11-08)_Utpal_12_Chemistry_2_12_html_m592d4422.gif

= 5.08 g (approx)

Hence, 5.08 g of ascorbic acid is needed to be dissolved.

Note: There is a slight variation in this Answer and the one given in the NCERT textbook.

Question 2.12:

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C.

Answer:

It is given that:

Volume of water, V = 450 mL = 0.45 L

Temperature, T = (37 + 273)K = 310 K

Number of moles of the polymer, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/261/6125/NCERT(INTEXT)_18-11-08)_Utpal_12_Chemistry_2_12_html_17dbeb5.gif

We know that:

Osmotic pressure, https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/261/6125/NCERT(INTEXT)_18-11-08)_Utpal_12_Chemistry_2_12_html_m76d47411.gif

https://img-nm.mnimgs.com/img/study_content/curr/1/12/17/261/6125/NCERT(INTEXT)_18-11-08)_Utpal_12_Chemistry_2_12_html_m10559035.gif

= 30.98 Pa

= 31 Pa (approximately)

Also Read : Page-No-92:-Chapter-3-Electrochemistry-Exercise-Solutions-class-12-ncert-solutions-Chemistry

SaraNextGen